Consider a system where a particle is placed in a uniform gravitational field $\vec{F} = -mg\,\vec{e}_{z}$. The dynamics of this are clearly invariant under translations. When we take $z\rightarrow z+a$ for any constant $a$, the force law stays the same, and the dynamics don't change at all.

The corresponding Lagrangian is $$ L = \frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}) - mgz. $$ The Lagrangian $L$ and the action $S = \int_{t_{0}}^{t_{1}} L\, dt$ both fail to be invariant under translations $z\rightarrow z + a$ (they end up differing by constants under this transformation).

Moreover, we find that the $z$-component of the momentum of the particle $p_{z} = m\dot{z}$ is not conserved. Under the gravitational force, the momentum is constantly gained downward at a rate $\dot{p}_{z} = -mg$.

Why is it that

  1. in the Newtonian framework, everything seems translation invariant, yet

  2. in the Lagrangian framework we don't seem to have translation invariance, and

  3. momentum is not conserved?

How are statements (1), (2), (3) reconciled with each other?


1 Answer 1

  1. A quasi-symmetry of the action is also a symmetry of the Euler-Lagrange (EL) equations; but the opposite does not necessarily hold, cf. e.g. this related Phys.SE post.

  2. OP's example is not a strict symmetry of the action but it is still a quasi-symmetry of the action. The corresponding full Noether charge $Q=p_z+mgt$ is hence conserved.

  • $\begingroup$ What is the prescription by which you get the conserved quantity corresponding to a quasi-symmetry? References? $\endgroup$ Jun 25 at 13:37
  • $\begingroup$ One method is the following trick. $\endgroup$
    – Qmechanic
    Jun 25 at 13:40

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